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Hardness of Approximating Multicut. S. Chawla, R. Krauthgamer, R. Kumar, Y. Rabani, D. Sivakumar (2005) Presented by Adin Rosenberg. Multicut. Input: An undirected graph G=(V,E), where |V|=n k pairs of vertices {s i ,t i } i=1,…,k , called demand pairs Optional: a cost function c on E
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Hardness of Approximating Multicut S. Chawla, R. Krauthgamer, R. Kumar, Y. Rabani, D. Sivakumar (2005) Presented by Adin Rosenberg
Multicut • Input: • An undirected graph G=(V,E), where |V|=n • k pairs of vertices {si,ti}i=1,…,k, called demand pairs • Optional: a cost function c on E • Goal: • A multicut: a subset of edges M, whose removal disconnects all of the demand pairs. • Of course, minimize c(M) (or |M| if c isn’t defined)
What are we going to prove? • Assuming the Unique Games conjecture is true, Multicut is NP-hard to approximate within any constant factor L How are we going to prove this? • We will show a reduction from a UG instance to a Multicut instance.
Unique Games • Input: • A bipartite graph G=(Q,EQ) • Each side p=1,2 contains n=|Q|/2 vertices (or questions) labeled q1p, q2p, …, qnp • Each edge (qi1,qj2) (called a question edge) is associated with a bijection bij:[d]→[d] • Each edge (qi1,qj2) has a nonnegative (normalized) weight wij
Unique Games (cont.) • A solution is an answer 1≤Aip≤d for each question qip • A solution satisfies an edge (qi1,qj2) if the answers Ai1 and Aj2 agree, i.e. Aj2=bij(Ai1) • Goal: • Find a solution with maximum value (total weight of satisfied edges)
Unique Games Conjecture [Khot 2002] • For every η,δ>0 there exists d=d(η,δ) such that it is NP-hard to determine whether a unique 2-prover game with answer set of size d has a value of: • at least (1- η), or • at most δ
A Little About Hypercubes • A d-dimensional hypercube is a graph G=(V,E) where V={0,1}d and there is an edge between two vertices if they differ in exactly one coordinate. • An edge (u,v) is called a dimension-a edge if u and v differ in coordinate a. • A dimension-a cut is the set of dimension-a edges. • The antipodal of a vertex u is the vertex which differs from u in every coordinate.
A Little About Hypercubes (0,1,1) (1,1,1) (0,0,1) (1,0,1) Dimension-1 edges (0,1,0) (1,1,0) (0,0,0) (1,0,0)
The Reduction from Unique Games to Multicut • For every vertex qip construct a d-dimensional hypercube Cip. Let the edges of these 2n cubes (called hypercube edges) have cost 1. • For each question edge (qi1,qj2) extend bij to a bijection b’ij:{0,1}d→{0,1}d defined by • Connect each vertex with using and edge (called a cross edge) with cost wijΛ, where Λ=n/η. • Define the demand pairs to be the antipodal pairs.
The Reduction from Unique Games to Multicut w11Λ w11 1 1 w23Λ 1 1 w23
The Yes Instance • Claim: If there is a solution A for the unique 2-prover game with value of at least 1-η, then there exists a multicut M for the Multicut instance such that c(M) ≤ 2d+1n • Proof: Construct the following multicut M: • For every answer Aip take the dimension- Aip cut in cube Cip. • For every edge (qi1,qj2) that the solution A doesn’t satisfy, take all the cross edges between Ci1 and Cj2.
The Yes Instance (cont.) • Removing M disconnects all the demand pairs: • For every vertex v in Cip, define f(v) to be the Aip-th coordinate of v. • For every edge (u,v) left, f(u)≠f(v) • The cost of M is at most 2d+1n: • Let S be the set of question edges not satisfied by the solution A.
A Little More About Hypercubes • For a function f on the vertices of a hypercube, define Iaf to be the fraction of dimension-a edges (u,v) for which f(u) ≠ f(v). • For a cutset M in a hypercube, define IaM to be the fraction of dimension-a edges that belong to M. • Observe that |M| = 2d-1ΣaIaM • And now some lemmas…
Lemma 1 • Let M be a cutset in a hypercube, and let g be the function labeling each vertex with the index of the connected component it belongs to. Then IaM≥Iag. • Proof: M contains every edge (u,v) for which g(u) ≠ g(v)
Lemma 2 • Let M be a cutset in a hypercube H. • Suppose M disconnects at least a β fraction of the antipodal pairs in H. • Then for every x>0, if ΣaIaM ≤ βx then there exists a dimension a* such that Ia*M ≥ 2-6x/27
Lemma 3 • Let f,g be two function on the vertices of a hypercube. • If f(v) = g(v) for all but a β fraction of the vertices v, then for every dimension a we have |Iaf – Iag| ≤ 2β.
The No Instance • Claim: There exists a constant c such that if the Multicut instance has a cutset of cost at most 2dnL (where L=c(log(1/(η+δ))) ) whose removal disconnects α≥7/8 fraction of the demand pairs, then there exists a solution for the unique 2-prover game whose value is larger than δ.
The No Instance (cont.) • Proof: • Let M we such a cutset for the Multicut instance. • Let Iap,i be the influence of M for each cube Cip. • Construct a randomized solution A for the unique 2-prover game instance. • For each vertex qip, we choose Aip to be the answer a with probability Iap,i / Σa Iap,i. • The expected value of A is at least δ, and therefore there exist a solution with such a value.
The No Instance (cont.) • Bound the probability of the following “bad” events (for a choice of the question edge (qi1,qj2) ): • E1 – fewer than half the demand pairs in Ci1 are disconnected in G \ M • E2 – M contains more than 2d+2L hypercube edges in Ci1. • E3 – M contains more than 2d+2L hypercube edges in Cj2. • E4 – M contains more than 2d / 296L+7 cross edges between Ci1 and Cj2. • All “bad” events do not occur with probability of at least 1/8.
The No Instance (cont.) • Assuming none of the “bad” events occur: • There exists a dimension a* s.t. Ia*1,I ≥ 2-96L/27 (according to Lemma 2) • Ibij(a*)2,j ≥ Ia*1,i – 2-96L-6 ≥ 2-96L/54 (Lemma 3) • The expected value of A is
What have we seen? • If the unique game has a value greater than 1-η, then the Multicut instance has a cutset M which disconnects all of the demand pairs with cost c(M) ≤ 2d+1n • If the unique game has a value less than δ, then the disconnected 7/8 of the demand pairs in the Multicut instance costs (at least) 2dLn • Therefore, the Unique Games Conjecture implies that it is NP-hard to approximate Multicut within a factor of any L>0.
Proof of Lemma 2 • ΣaIaM ≤ βx means Ia*M ≥ 2-6x/27 for some a* • Proof: • Convert M to a two-sided cut M’ • Define a Boolean function f according to the connected components of M’ • Use KKL’s lemma: ΣaIaf/α + Σa(Iaf)4/3 ≥ 2p*logα/α (where f is a Boolean function on a hypercube and p ≤ ½ is the balance of f)
Proof of Lemma 3 • If f(v) = g(v) for all but a β fraction of the vertices v, then for every dimension a we have |Iaf – Iag| ≤ 2β. • Proof: • For all but at most β2d of the edges, we have f(u)=g(u) and f(v)=g(v). • Therefore, only a β2d/2d-1=2β fraction of edges can contribute to the difference between Iaf and Iag.
Credits to… • The authors of the paper for giving me what to talk about. • Kahn, Kalai and Linial for saving us the Fourier analysis. • Sarai for taking care of the lights. • And of course, thank you all for listening…